The LR circuit

A coil that holds the current back

Close a switch in a circuit with a plain lamp, and the lamp lights at once. Put a big coil in series with the lamp, and the lamp takes a moment to reach its full brightness. Open the switch of a circuit with a big coil, and a spark may jump across the switch.

In both cases the coil is holding back a change in the current. Let us find how much, and for how long.

We need one fact about coils. When the current in a coil changes, the flux through the coil changes too. Faraday's law then gives the coil an emf, $-L\dfrac{dI}{dt}$. The number $L$ is the inductance of the coil, in henry (H). By Lenz's law the emf always opposes the change of the current.

Switching on

A battery of emf $\mathcal{E}$, a resistor $R$ and a coil of inductance $L$ are joined in series. The resistance of the coil's own wire is part of $R$. We close the switch at $t = 0$. The sum of the emfs and the voltage drops round the loop is zero:

Look at the two ends of the story.

• At the start. The current is zero, so $IR = 0$. The whole emf of the battery appears across the coil, and $\dfrac{dI}{dt} = \dfrac{\mathcal{E}}{L}$. The coil acts like a break in the circuit.
• A long time later. The current has stopped changing, so $L\dfrac{dI}{dt} = 0$. Then $I = \dfrac{\mathcal{E}}{R}$. The coil acts like a plain piece of wire.

In between, the current grows smoothly from the one to the other:

The quantity $\tau$ is the time constant. Its unit is a second, because $\text{H} = \text{V s/A}$ and $\Omega = \text{V/A}$, so $\text{H}/\Omega = \text{s}$.

After one time constant the current has reached $63\ \%$ of its final value. After two, $86\ \%$. After three, $95\ \%$. After five, more than $99\ \%$. A larger coil, or a smaller resistance, makes a longer time constant and a slower rise.

The voltage across the coil is $L\dfrac{dI}{dt} = \mathcal{E}e^{-t/\tau}$. It starts at the full $\mathcal{E}$ and falls away. The voltage across the resistor is $IR = \mathcal{E}\left(1 - e^{-t/\tau}\right)$. The two always add up to $\mathcal{E}$.

To find the time to reach a fraction $f$ of the final current, set $1 - e^{-t/\tau} = f$. We want $t$, so $t = -\tau\ln(1 - f)$. For the half-way point, $t = \tau\ln 2 = 0.693\,\tau$.

Example: switching on

A coil is in series with a battery of $12$ V, and the total resistance is $8.0\ \Omega$. The inductance is $1.2$ H. Find the time constant, the final current, and the current $0.30$ s after the switch is closed.

1. The time constant. $\tau = \dfrac{L}{R} = \dfrac{1.2\ \text{H}}{8.0\ \Omega} = 0.15\ \text{s}$.
2. The final current. $\dfrac{\mathcal{E}}{R} = \dfrac{12\ \text{V}}{8.0\ \Omega} = 1.5\ \text{A}$.
3. The starting rate. $\dfrac{dI}{dt} = \dfrac{\mathcal{E}}{L} = \dfrac{12\ \text{V}}{1.2\ \text{H}} = 10\ \text{A/s}$.
4. At $0.30$ s. That is $\dfrac{0.30}{0.15} = 2$ time constants. $I = 1.5\ \text{A} \times \left(1 - e^{-2}\right) = 1.5\ \text{A} \times 0.865 = 1.30\ \text{A}$.
5. Across the coil then. $\mathcal{E}e^{-2} = 12\ \text{V} \times 0.135 = 1.6\ \text{V}$.

Try the smallest and the largest inductance. How many time constants does the current need to get within a few per cent of its final value? The answer is the same every time, but the time in ms changes.

Switching off

Now suppose the coil carries a current $I_0$ and we cut the battery out, but give the current a path through a resistor, so that the total resistance is $R$. Round the loop, $-L\dfrac{dI}{dt} = IR$. The current dies away:

It falls to $37\ \%$ after one time constant. The coil keeps the current flowing for a while, using the energy in its magnetic field. All of the energy, $\tfrac{1}{2}LI_0^2$, goes into heat in the resistor. We can check this: the heat is $\int I^2R\,dt = I_0^2R \times \dfrac{\tau}{2} = \dfrac{1}{2}LI_0^2$.

To find the time for the current to fall by a given fraction, solve $e^{-t/\tau} = 1 - p$ the same way.

For the coil of the last example, joined to a path of $8.0\ \Omega$ at the moment the battery is cut out, the current starts at $1.5$ A. After $0.10$ s it is $1.5 \times e^{-0.10/0.15} = 0.77$ A. The heat that comes out in all is $\tfrac{1}{2} \times 1.2\ \text{H} \times (1.5\ \text{A})^2 = 1.35$ J.

The spark at the switch

What if we simply open the switch, so that there is no path at all? The current has to fall almost to zero in a very short time. Then $\dfrac{dI}{dt}$ is very large, and so is the emf, $L\left|\dfrac{\Delta I}{\Delta t}\right|$.

A coil of $2.5$ H that carries $0.80$ A, with the current broken in $0.50$ ms, has an average emf of

This is enough to make a spark jump across the gap of the switch, though the battery is only a few volts. The emf is in the direction that tries to keep the current going. It is why a relay in a circuit has a diode or a resistor across its coil, to give the current a safe path.

Practise with new numbers

Try it in Python

The program finds the current in a coil at one, two, three, four and five time constants. Then it follows the same current in tiny steps of time, from the rule $L\,dI/dt = \mathcal{E} - RI$.

Common mistakes

• Thinking the current rises at once to $\mathcal{E}/R$. The coil holds the current back at first. The current is zero at $t = 0$.
• Taking $\tau$ as the time to finish. After one time constant the current has done only $63\ \%$ of its rise. It needs about five to finish.
• Mixing the units of $\tau$ and $t$. Use seconds for both, or ms for both.
• Using $\tau = LR$. The time constant is $L/R$.
• Thinking the coil has no effect when the current is steady. It has no emf then, but it still holds energy $\tfrac{1}{2}LI^2$.
• Forgetting the resistance of the coil's own wire. It is part of $R$.
• Thinking the spark comes from the battery. The large emf comes from the coil, from the rapid fall of its current.

Remember these

• The emf of a coil is $-L\dfrac{dI}{dt}$, and it opposes the change of the current.
• Switching on: $I = \dfrac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)$. The voltage across the coil is $\mathcal{E}e^{-t/\tau}$.
• Switching off through a resistor: $I = I_0e^{-t/\tau}$.
• The time constant is $\tau = \dfrac{L}{R}$, in seconds. After $\tau$: $63\ \%$ done, or $37\ \%$ left.
• At the start the coil acts like a break in the circuit. After a long time it acts like a wire.
• Time to a fraction $f$ of the final current: $t = -\tau\ln(1 - f)$.
• Breaking a current at once gives $L\Delta I/\Delta t$, a very large emf and a spark.
• All of $\tfrac{1}{2}LI_0^2$ is given back as heat in the resistor.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A coil of inductance $0.60$ H and resistance $3.0\ \Omega$ is joined to a battery. What is the time constant?

2. A coil and a resistor are joined in series to a battery. At the instant the switch is closed, what is the current, and what is the voltage across the coil?

3. The same circuit has been on for a long time. What are the current and the voltage across the coil now?

4. Why can a spark jump across a switch when it opens a circuit that has a large coil in it?

5. The resistance in an LR circuit is halved, and nothing else changes. What happens to the time constant, and to the final current?